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Algebra / Systems of two linear equations in two variables Difficulty: Hard

$48 x - 72 y = 30 y + 24$

$r y = \frac{1}{6} - 16 x$

In the given system of equations, $r$ is a constant. If the system has no solution, what is the value of $r$ ?

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Explanation

The correct answer is $negative 34$ . A system of two linear equations in two variables, $x$ and $y$ , has no solution if the lines represented by the equations in the xy-plane are distinct and parallel. Two lines represented by equations in standard form $A x + B y = C$ , where $A$ , $B$ , and $C$ are constants, are parallel if the coefficients for $x$ and $y$ in one equation are proportional to the corresponding coefficients in the other equation. The first equation in the given system can be written in standard form by subtracting $30 y$ from both sides of the equation to yield $48 x - 102 y = 24$ . The second equation in the given system can be written in standard form by adding $16 x$ to both sides of the equation to yield $16 x + r y = \frac{1}{6}$ . The coefficient of $x$ in this second equation, $16$ , is $\frac{1}{3}$ times the coefficient of $x$ in the first equation, $48$ . For the lines to be parallel the coefficient of $y$ in the second equation, $r$ , must also be $\frac{1}{3}$ times the coefficient of $y$ in the first equation, $negative 102$ . Thus, $r = \frac{1}{3} (- 102)$ , or $r = - 34$ . Therefore, if the given system has no solution, the value of $r$ is $negative 34$ .